Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $n \neq 0$. $y = \dfrac{2n - 4}{-n^2 + 6n + 16} \times \dfrac{n^2 - 11n + 24}{5n - 15} $
First factor out any common factors. $y = \dfrac{2(n - 2)}{-(n^2 - 6n - 16)} \times \dfrac{n^2 - 11n + 24}{5(n - 3)} $ Then factor the quadratic expressions. $y = \dfrac {2(n - 2)} {-(n - 8)(n + 2)} \times \dfrac {(n - 8)(n - 3)} {5(n - 3)} $ Then multiply the two numerators and multiply the two denominators. $y = \dfrac {2(n - 2) \times (n - 8)(n - 3) } { -(n - 8)(n + 2) \times 5(n - 3)} $ $y = \dfrac {2(n - 8)(n - 3)(n - 2)} {-5(n - 8)(n + 2)(n - 3)} $ Notice that $(n - 8)$ and $(n - 3)$ appear in both the numerator and denominator so we can cancel them. $y = \dfrac {2\cancel{(n - 8)}(n - 3)(n - 2)} {-5\cancel{(n - 8)}(n + 2)(n - 3)} $ We are dividing by $n - 8$ , so $n - 8 \neq 0$ Therefore, $n \neq 8$ $y = \dfrac {2\cancel{(n - 8)}\cancel{(n - 3)}(n - 2)} {-5\cancel{(n - 8)}(n + 2)\cancel{(n - 3)}} $ We are dividing by $n - 3$ , so $n - 3 \neq 0$ Therefore, $n \neq 3$ $y = \dfrac {2(n - 2)} {-5(n + 2)} $ $ y = \dfrac{-2(n - 2)}{5(n + 2)}; n \neq 8; n \neq 3 $